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# sum of the first four numbers in a list

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Director
Joined: 21 Dec 2009
Posts: 592
Concentration: Entrepreneurship, Finance
Followers: 16

Kudos [?]: 333 [0], given: 20

sum of the first four numbers in a list [#permalink]  04 Aug 2010, 10:03
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (01:12) wrong based on 5 sessions
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?
_________________

KUDOS me if you feel my contribution has helped you.

Math Expert
Joined: 02 Sep 2009
Posts: 27170
Followers: 4226

Kudos [?]: 40953 [1] , given: 5576

Re: sum of the first four numbers in a list [#permalink]  04 Aug 2010, 10:31
1
KUDOS
Expert's post
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

The way you are doing is also valid. You've just made an error in calculations, plus no need even number to be $$2x$$ it can be just even $$x$$.

Sum of the first four: $$(2x-4)+(2x-2)+(2x)+(2x+2)=8x-4=908$$;
Sum of the last four: $$2x+(2x+2)+(2x+4)+(2x+6)=8x+12=(8x-4)+16=908+16=924$$.

Hope it's clear.
_________________
Director
Joined: 21 Dec 2009
Posts: 592
Concentration: Entrepreneurship, Finance
Followers: 16

Kudos [?]: 333 [0], given: 20

Re: sum of the first four numbers in a list [#permalink]  04 Aug 2010, 11:11
oh, thanks Bunuel for the corrections.
Hope you aren't tired of receiving kudos; you know, people like you have made
kudos a trite, and one simply wonders what else to give.

Methinks, there should be a different category of kudos for genius such as Bunuel.
_________________

KUDOS me if you feel my contribution has helped you.

Senior Manager
Joined: 23 May 2010
Posts: 442
Followers: 5

Kudos [?]: 41 [0], given: 112

Re: sum of the first four numbers in a list [#permalink]  07 Aug 2010, 03:44
Bunuel wrote:
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx
Math Expert
Joined: 02 Sep 2009
Posts: 27170
Followers: 4226

Kudos [?]: 40953 [0], given: 5576

Re: sum of the first four numbers in a list [#permalink]  07 Aug 2010, 04:04
Expert's post
gauravnagpal wrote:
Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx

$$4x+ 12 =908$$ --> $$x=224=even$$.
_________________
Manager
Joined: 29 Jul 2010
Posts: 127
Followers: 1

Kudos [?]: 3 [0], given: 47

Re: sum of the first four numbers in a list [#permalink]  07 Aug 2010, 05:38
Bunuel wrote:
gmatbull wrote:
If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
A. 912
B. 914
C. 916
D. 920
E. 924

I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
--> 4x-4=908
x=228

sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
= 8x+12
=8(228) + 12

Please what is the correct approach?

Let the six consecutive even numbers be $$x$$, $$x+2$$, $$x+4$$, $$x+6$$, $$x+8$$, $$x+10$$.

Given: $$x+(x+2)+(x+4)+(x+6)=4x+12=908$$. Question: $$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=?$$

$$(x+4)+(x+6)+(x+8)+(x+10)=4x+28=(4x+12)+16=908+16=924$$.

The way you are doing is also valid. You've just made an error in calculations, plus no need even number to be $$2x$$ it can be just even $$x$$.

Sum of the first four: $$(2x-4)+(2x-2)+(2x)+(2x+2)=8x-4=908$$;
Sum of the last four: $$2x+(2x+2)+(2x+4)+(2x+6)=8x+12=(8x-4)+16=908+16=924$$.

Hope it's clear.

Well done!!!!

Thx and kudos
Senior Manager
Joined: 23 May 2010
Posts: 442
Followers: 5

Kudos [?]: 41 [0], given: 112

Re: sum of the first four numbers in a list [#permalink]  08 Aug 2010, 00:49
Bunuel wrote:
gauravnagpal wrote:
Hi Bunuel
i did the same way ..
however kindly explain this if I solve
4x+ 12 =908 , I get x = 99 ..which is not an even number . hence I was confuded whether I missed something !!
thanx

$$4x+ 12 =908$$ --> $$x=224=even$$.

i am so sorry ...i reckon sleep of mind ...i dont know how could I write this ..thanx anyways for spending time on this
Re: sum of the first four numbers in a list   [#permalink] 08 Aug 2010, 00:49
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